What is the derivative of f(x)= 1/(1+e^{-x)} ?

The derivative of f(x)= 1/(1+e^{-x)} is (e^{-x})/((1+e^{-x))^2}

What is the first derivative of f(x)= 1/(1+e^{-x)} ?

The first derivative of f(x)= 1/(1+e^{-x)} is (e^{-x})/((1+e^{-x))^2}

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derivative of

f (x) = \frac{1}{1 + e ^{- x}}

\frac{e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1}{1 + e ^{- x}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{d}{d x} ((1 + e^{- x})^{- 1})

Apply the chain rule:

- \frac{1}{( 1 + e ^{- x} ) ^{2}} \frac{d}{d x} (1 + e^{- x})

= - \frac{1}{( 1 + e ^{- x} ) ^{2}} \frac{d}{d x} (1 + e^{- x})

\frac{d}{d x} (1 + e^{- x}) = - e^{- x}

= - \frac{1}{( 1 + e ^{- x} ) ^{2}} (- e^{- x})

Simplify

- \frac{1}{( 1 + e ^{- x} ) ^{2}} (- e^{- x}) : \frac{e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

= \frac{e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

\frac{d}{d x} ((x + 1) e^{x^{2} - 4})

\int cos (\frac{π}{2} x) d x

\frac{d}{d x} ((4 x^{2} - x) (x^{3} - 8 x^{2} + 12))

f (x) = x^{2} + 3

x \to 1 lim (4 x - \frac{4}{x - 1})

What is the derivative of f(x)= 1/(1+e^{-x)} ?
The derivative of f(x)= 1/(1+e^{-x)} is (e^{-x})/((1+e^{-x))^2}
What is the first derivative of f(x)= 1/(1+e^{-x)} ?
The first derivative of f(x)= 1/(1+e^{-x)} is (e^{-x})/((1+e^{-x))^2}